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Evgen [1.6K]
3 years ago
8

And wins a metal in the shape of a circle for finishing a marathon which statements about the metals are true use 3.14 for pi. S

elect all that are correct A. If the Metal has a circumference of seven pi inches Then it has a diameter Of 3.5in. B. If the metal has a circumference of three pi inches then it has an area of nine pi square inches. C. If the metal has a radius of 2.5 inches then it has an area of 19.625 in.² D. If the metal has a diameter of 3 inches and it has a circumference of 9.42 inches. E. If the metal has a circumference of four pi inches then it has an area of four pi square inches
Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
8 0

Answer:

The true statements are:

C. If the metal has a radius of 2.5 inches then it has an area of 19.625 in.²

D. If the metal has a diameter of 3 inches and it has a circumference of 9.42 inches.

E. If the metal has a circumference of four pi inches then it has an area of four pi square inches

Step-by-step explanation:

Let us revise the rule of circumference and area of a circle

  • The formula of the circumference of a circle is <em>C = π d</em>, where d is the diameter of the circle OR <em>C = 2πr</em>, where r is the radius of the circle
  • The formula of the area of a circle is <em>A = πr²</em>
  • The radius of a circle is equal to half its diameter

A.

∵ The metal has a circumference of 7π inches

∵ C = 2πr

- Equate πd by 7π

∴ πd = 7π

- Divide both sides by π

∴ d = 7 inches not 3.5 inches

∴ A is not true

B.

∵ The metal has a circumference of 3π inches

∵ C = 2πr

- Equate 2πr by 3π

∴ 2πr = 3π

- Divide both sides by π

∴ 2r = 3

- Divide both sides by 2

∴ r = 1.5 inches

∵ A = πr²

∴ A = π(1.5)²

∴ A = 2.25π inches² not 9π inches²

∴ B is not true

C.

∵ The metal has a radius of 2.5 inches

∴ r = 2.5

∵ A = πr²

∵ π = 3.14

∴ A = 3.14(2.5)²

∴ A = 19.625 inches²

∴ If the metal has a radius of 2.5 inches, then its area is 19.625 in.²

∴ C is true

D.

∵ The metal has a diameter of 3 inches

∴ d = 3

∵ C = πd

∵ π = 3.14

∴ C = 3.14(3)

∴ C = 9.42 inches

∴ If the metal has a diameter of 3 inches, then its circumference

  is 9.42 inches

∴ D is true

E.

∵ The metal has a circumference of 4π inches

∵ C = 2πr

- Equate 2πr by 4π

∴ 2πr = 4π

- Divide both sides by 2π

∴ r = 2

- Now you can find the area

∵ A = πr²

∴ A = π(2)²

∴ A = 4π

∴ If the metal has a circumference of 4π inches, then its area

   is 4π  inches²

∴ E is true

The true statements are:

C. If the metal has a radius of 2.5 inches then it has an area of 19.625 in.²

D. If the metal has a diameter of 3 inches and it has a circumference of 9.42 inches.

E. If the metal has a circumference of four pi inches then it has an area of four pi square inches

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Step-by-step explanation:

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<h3><u>Answer:</u></h3>

\large\boxed{\pink{\sf \leadsto Value \ of \ LM \ is \ 11 \ units . }}

<h3><u>Step-by-step explanation:</u></h3>

A figure is given to us in which we can see two triangles one is ∆ MPL and other is ∆MPN .

<u>Figure</u><u> </u><u>:</u><u>-</u><u> </u>

\setlength{\unitlength}{1 cm}\begin{picture}(12,12)\linethickness{0.25mm}\put(0,0){\line(1,2){2}} \put(0.001,0){\line(1,0){4}}\put(2,0){\vector(0,1){5}}\put(4,0){\line( - 1,2){2}}\put(0,-0.4){$\bf L $}\put(2,-0.4){$\bf P$}\put(4,-0.4){$\bf N $}\put(2.2,4){$\bf M $}\put(2.8, - .4){$\bf 5 $}\put(1, - .4){$\bf 5 $}\put(3.4, 2){$\bf 11 $}\put(2.3,0){\line(0,1){.3}}\put(2.3,.3){\line( - 1,0){.3}}\end{picture}

\underline{\blue{\sf In\: \triangle MPL \ \& \ \triangle MPN :- }}

\qquad \bullet LP = PN = 5 \:\:(given) \\\\\qquad \bullet MP = MP \:\:(Common) \\\\\qquad \bullet \angle MPN = \angle MPL = 90^{\circ} \:\: (given)

Hence by SAS congruence condition ,

\orange{\bf \triangle MPL \cong \triangle MPN }

Hence by cpct ( Corresponding parts of congruent triangles ) we can say that , LM = NM = 11 units .

<h3><u>Hence </u><u>the</u><u> </u><u>value</u><u> </u><u>of</u><u> </u><u>LM</u><u> </u><u>is</u><u> </u><u>1</u><u>1</u><u> </u><u>units</u><u> </u><u>.</u></h3>
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